sag amount vs. No. of turns on ring nut?

I had no luck with my search on this topic even though I remember reading it maybe a year ago or so...

Does anybody have a rule of thumb on how many turns it takes on the spring pre-load ring nut that will give you about 5mm of race sag?

I don't know if it is the same for the WR250F, which is what I have. But one turn = 1.5mm so you would be about 3 1/3 turns.

better get confirmation from a big bore though.

mook

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2001 WR250F

1975 KZ900 - project bike

I don't think 3.5 turns is right. That's the shock travel difference, not the sag difference (5.5" of shock travel=over 12" of suspension travel)

I've found that slight movements in the rings make a BIG difference here. Mark where you are with a crayon so we can guage how far you turned the rings. Try just 3 "knotches" first and measure...IMHO

I would think that the number of turns of the ring nut and resulting sag would be dependent on the weight of rider and gear. What do I know though?...I work in a grocery store.

My .02

Good Luck,

Bryce

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Bryce Senff

WMRRA #201

WR426F

TL1000SV

CBR900RR

Oh it's Good to Be Back on a Dirtbike!

Gotta go with Bryce on this one...race sag will be determined by rider weight w/ gear...

free sag will be determined by how worked over the rear spring is(even fresh springs will have a slight variations if you throw them on a spring tester)....no way to determine these settings by # of turns unless you were consistent on the rest of the variables in the equation. Turn and measure is still the accurate way to set sag.

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